Free Essay

Transfer Time

In: Computers and Technology

Submitted By Dacrazefury
Words 589
Pages 3
Brian Rodriguez
Professor David Durniok
Assignment 2

Transfer Time

1. Fill in the Transfer Time column in Table 1. State all times in appropriate units (seconds, minutes, or hours). Show all your calculations and number conversions.

File Size | Transfer Rate | Transfer Time | 100 MB | 56 Kbps | 5 hrs | 100 MB | 4 Mbps | 3.3 minutes | 2 GB | 56 Kbps | 83.22 hours | 2 GB | 4 Mbps | 1 hr, 8 mins, 16 seconds |

* Take the 56kbps = 56,000 BITS per second, divide by 10 and you get 5,600 bytes. 100,000,000 / 5,600 = 17,857 seconds = 297.6 minutes = 4.96 hrs (almost 5 hrs) if there is no interruption in the transmittal.

* 4Mbps (megabits) equals to approximately 500KB/s (kilobytes).

100MB (Mega Bytes) equals approximately 100000 KB

So assuming you can have a perfect transfer rate of 500KB/sec (assuming there is no TCP/IP overhead) you divide 100000KB by your 500KB/Sec transfer rate:

= 200 seconds

200 seconds divided by 60 = 3.3 minutes.

* 2GB * 1024MB/GB = 2048MB
2048MB * 1024KB/MB = 2097152 KB
2097152KB * 8Kb/KB = 16777216 Kb
16777216Kb / 56Kb/s = 299593s
299593s / 3600s/h = 83.22h

so, at the theoretical rate of 56Kb/s, it would take 83.22 hours do download a 2GB file. At a more realistic average download rate of 4Kb/s, it would take 322.88 hours or just under two weeks to download.

* 4 Mbps = 0.5 MB/s

The transfer rate is 0.5 MB/s

2048 / 0.5 = 4096 seconds

Time taken = 1 hour 8 minutes 16 seconds

2. Fill in the File Size column in Table 2. State all sizes in appropriate units (KB, MB, or GB). Show all your calculations and number conversions.

Transfer Rate | Transfer Time | File Size | 56 Kbps | 30 minutes | 12.6MB | 4 Mbps | 10 minutes | 2.4MB | 56 Kbps | 4 hours | 785.7MB | 4 Mbps | 2 hours | 4800 MB |

* 56000bits/second = 56000/8 bytes per second for 30 minutes, effectively it should be able to download, 7kbytes * 1800
Hence, the file size is 12.6MB or

* A file at a rate of 4 Mbps (Megabits per second) for a period of 10 minutes.

You would multiply the transfer rate (4 Mbps) by 60 (the number of seconds in a minute), then multiply that by 10 (the number of minutes it's taking).

* File size will be
56*3600(One hour has 3600 seconds)*4
806400 kb = 785.7 MB

* 4 Mbps == 4Mb * 60 * 10 = 2400 Mb (Megabytes)

3. Comment on the relationship among file size, transfer rate, and transfer time. How are they related?

* If the transfer connection and software is decent, then the relationship should be as follows: Transfer rate is constant, file size is arbitrary, and transfer time is proportional to file size and the reciprocal of transfer rate.

If we call the transfer rate R, the transfer time T, and the file size S, we have:

T = S/R

We can rearrange this to get:

S = R*T
R = S/T

Citation: Yahoo Answer/Toms Hardware

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